什么是C ++中的运算符T *(其中T是模板参数)? [英] What is operator T* (where T is a template parameter ) in C++?

问题描述

class NullClass{
    public:
    template<class T>
        operator T*() const {return 0;}

};

我正在阅读有效的C ++，遇到了这个类，我实现了该类并进行编译.我对此有一些疑问:

I was reading Effective C++ and I came across this class, I implemented the class and it compiles. I have a few doubts over this:

1. 它没有返回类型.

1. It doesn't have a return type.

这是什么运算符.

及其实际功能.

推荐答案

这是类型转换运算符.它定义了类的实例和指定的类型(此处为T*)之间的隐式转换.它的隐式返回类型当然是相同的.

That's the type conversion operator. It defines an implicit conversion between an instance of the class and the specified type (here T*). Its implicit return type is of course the same.

这里NullClass实例，当提示转换为任何指针类型时，将产生从0到所述类型的隐式转换，即该类型的空指针.

Here a NullClass instance, when prompted to convert to any pointer type, will yield the implicit conversion from 0 to said type, i.e. the null pointer for that type.

顺便说一句，可以明确指定转换运算符:

On a side note, conversion operators can be made explicit :

template<class T>
explicit operator T*() const {return 0;}

这避免了隐式转换(它可能是错误的细微来源)，但允许使用static_cast.

This avoid implicit conversions (which can be a subtle source of bugs), but permits the usage of static_cast.

这篇关于什么是C ++中的运算符T *(其中T是模板参数)?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！